How to Solve the Non-linear Differential Equation in Radial Ink Diffusion?

[tanaygupta2000]

Homework Statement

The concentration p(r,t) of ink diffusing in water is governed by the diffusion equation
∂p/∂t = D∇^2(p)
where D is a parameter known as diffusion constant. What is the average time taken for a molecule of ink to spread by a root mean square distance R?
(a) √(R/D)
(b) R/√D
(c) R^2 /D
(d) RD

Relevant Equations

∂p/∂t = D∇^2(p)

I assumed p(r,t) as p(r,t) = R(r)T(t) as Separation of Variables method. I got the expression of T(t) as
T(t) = C₁e^C₂t

and got a non-linear differential equation in R(r) as
d²R/dr² + (2/r)dR/dr - (C/D)R = 0

(I assumed r to be the radial distance in spherical coordinates)
Now I'm not getting how to solve this differential equation.

 

[kuruman]

You don't need to solve a differential equation to answer this multiple choice question. Only one answer passes the test of dimensional analysis.

 

[tanaygupta2000]

You don't need to solve a differential equation to answer this multiple choice question. Only one answer passes the test of dimensional analysis.

R has the dimensions of length and D is a dimensionless quantity.
None of the options give the dimensions of time.

 

[kuruman]

D is not dimensionless. What are its dimensions as per the diffusion equation you posted?

 

[tanaygupta2000]

D is not dimensionless. What are its dimensions as per the diffusion equation you posted?

I'm assuming p(r,t) = R(r)T(t).
By this assumption, ∂p/∂t has the dimensions of [LT]/[T] = [L]
and ∇²p(r,t) has the dimensions of [LT]/[L²] = [L^-1T]

Hence only R²/D has the dimensions of time, hence the correct answer !

 

[kuruman]

I'm assuming p(r,t) = R(r)T(t).
By this assumption, ∂p/∂t has the dimensions of [LT]/[T] = [L]
and ∇²p(r,t) has the dimensions of [LT]/[L²] = [L^-1T]

Hence only R²/D has the dimensions of time, hence the correct answer !

Yes.